# If T unitary show H hermitian (and the reverse)

1. Apr 8, 2015

### ognik

1. The problem statement, all variables and given/known data
An operator T(t+ε,t) describes the change in the wave function from to to t+ε. For ε real and small enough so that ε2 may be neglected, considering the eqtn below:
(a)If T is unitary, show H is hermitian
(b)if H hermitian, show T is unitary.

2. Relevant equations
$$T(t+ε,t)= 1-\frac{i}{\hbar} \epsilon H(t)$$

3. The attempt at a solution
I've been making decent progress through a bunch of hermitian/unitary exercises, but this looks nothing them - or anything in the text, which have been all about matrix operation without any functions... I think firstly there might be some ways to work with equations using U and H matrices, if so a link or 2 or a beginners explanation would be great. Then also a hint as to how I might start this? Thanks :-)

2. Apr 8, 2015

### robphy

What are the definitions of being hermitian? unitary?

3. Apr 9, 2015

### ognik

Thanks for the interest Robphy :-) I tried (a) If T unitary :
$$T T^{\dagger} = 1 = (1-\frac{i}{\hbar}\epsilon H)(1-\frac{i}{\hbar}\epsilon H)^{\dagger}$$
$$= (1-\frac{i}{\hbar}\epsilon H)(1+\frac{i}{\hbar}\epsilon H^{\dagger})$$
But I've got something wrong? I cant work that so that $$H=H^\dagger$$
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Trying (b), solve for H:
$$H=-\frac{(T-1)}{\epsilon}\frac{\hbar}{i} = i(T-1)\frac{\hbar}{\epsilon}$$
If H hermitian, then
$$H^\dagger =- i(T^\dagger-1)\frac{\hbar}{\epsilon}=i(T-1)\frac{\hbar}{\epsilon}$$
$$\therefore -(T^\dagger-1)=(T-1)$$
which is also wrong. Must be in the way I found the hermitian of expressions, but I split that into conjugate first, then transpose - can't see what I'm doing wrong (after many pages of scrap paper). Please check my working and let me know what I've misunderstood?

4. Apr 9, 2015

### robphy

For (a), carry out the multiplication.

For (b), I don't think it's a good idea to solve for H since (as you now hopefully saw in (a)) you need to use the approximation.
Assume the condition on H, what does it imply for T? Is the condition needed for T satisfied?

Last edited: Apr 9, 2015
5. Apr 9, 2015

### ognik

Thanks - good to know my arithmetic was ok, then for (a) I got
$$1-\frac{i}{\hbar} \epsilon H + \frac{i}{\hbar}\epsilon H^\dagger - \frac{-1}{\hbar^2} {\epsilon}^2HH^\dagger$$
Even though the problem says to ignore the ε2 term, it would be multiplied by H2; also h bar is ~ 10-34, so while I could ignore the ε2 term, I can't comfortably see how to justify ignoring the whole of
$$\frac{1}{\hbar^2} {\epsilon}^2HH^2$$

I know the limit as ε→0 = 0, but (h bar)2 is pretty small also; I have no feel for how small ε can be in practice..... or how large H2 can be?
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Similarly for (B), without solving for H - and having worked through all the hermitian properties I know of, the only thing that looked promising was to evaluate TT*, which gave a similar expansion to the above. Then if H*=H when H hermitian (which it doesn't) .... then TT* also =1 if unitary ....
So, sorry if I'm being dense, but another hint at (b) please?
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I have finished all the other exercises in this section, except another similar one - so excuse me for referencing 2 in 1, but its probably the same bit of understanding missing in both; so thought it might be useful for any helpers to know that .....
The other eqtn is U=exp(iaH), a real. I easily showed U unitary if H hermitian, but again can't see how to do the reverse (I definitely don't want to solve for H here ;-) ) Thanks again.

Last edited: Apr 9, 2015
6. Apr 9, 2015

### Dick

You shouldn't worry about ignoring terms like $\epsilon^2$, it's an infinitesimal. It's supposed to be much smaller than anything else around regardless of the other values (i.e. it's infinitesimal). Just ignore it. Then you've got, $TT^\dagger=1-\frac{i}{\hbar} \epsilon (H - H^\dagger)$. If $T$ is unitary then $TT^\dagger=1$, what does that tell you about $H-H^\dagger$? If $H$ is Hermitian what does that tell you about $TT^\dagger$ and $T$?

And it is true that showing $U=exp(iaH)$ is unitary implies $H$ is Hermitian is a bit trickier than the reverse. In particular, you don't know a priori that $H$ and $H^\dagger$ even commute. So be careful. Start from $UU^\dagger=1$ and differentiate both sides with respect to a. See where that leads you.

Last edited: Apr 9, 2015
7. Apr 10, 2015

### Fredrik

Staff Emeritus
I don't think "we can ignore the $\varepsilon^2$ term because it's small" is the best argument here. The equality holds for all $\varepsilon$ in some small interval that contains 0. I would say that the statement that gives us the best reason to ignore the $\varepsilon^2$ isn't the "small", it's the "for all".

Consider the statement that a,b,c,r are real numbers such that $a+bx+cx^2=0$ for all real numbers x in the interval (-r,r). It implies that a=b=c=0. Similarly, if $a+bx+cx^2=d+ex+fx^2$ for all x, then the coefficients of each $x^n$ must be the same on both sides, so we have a=d, b=e, c=f.

If we apply this idea to the result that
$$1=1+\frac{i}{\hbar}\varepsilon(H^\dagger-H)+\frac{\varepsilon^2}{\hbar}H^\dagger H$$ for all $\varepsilon$ in some interval that contains 0, then the conclusion is that $H^\dagger=H$. At first it may seem that we can also conclude that $H^\dagger H=0$, but this would be wrong. The reason is the calculation has already neglected other terms of order $\varepsilon^2$, so the right-hand side is already wrong about second and higher order terms.

8. Apr 10, 2015

### ognik

All done, many thanks - was much easier than it first seemed once I stopped being too impressed by the actual equations :-)

Could you also - perhaps & please - have a look at https://www.physicsforums.com/threa...te-systems-scale-factors.806607/#post-5064384 and see if you can shed some light? Its from a previous section of the book, but I prefer to finish everything - its bothering me that I got close but seem to be missing something in terms of index and summation....Thanks again.

9. Apr 10, 2015

### Fredrik

Staff Emeritus
There's something missing in your problem statement in that thread. You haven't defined the symbols and the notations you use. Am I right to assume that boldface symbols denote elements of $\mathbb R^3$ and that when you put a hat on such a symbol, it denotes the corresponding unit vector (e.g. $\hat{\mathbf q}=\mathbf q/|\mathbf q|$)? You're mixing your notations in a weird way. Is there a difference between $\hat{\mathbf q}_i$ and $\hat q_i$? Is the "relevant equation" supposed to involve $\mathbf r$ or $\hat{\mathbf r}$?

It's very likely that you didn't get a reply because no one understood the question. If you explain the problem statement (in the other thread), I can take another look at it.

10. Apr 10, 2015

### BvU

Second Fredrik here. I had a look at that one too and couldn't make out heads or tails from the postings. A bit more context please.
My respect for your studying by correspondence and kudos for your tenacity. Keep it up and we'll help as best we can.

11. Apr 10, 2015

### ognik

Thanks Frederiks :-) My last post must have crossed with yours, I appreciate the extra on the infinitesimals (and wouldn't have concluded HH =0). I have not really been happy with the way infinitesimals are sometimes used, I will continue to thing about them as I blunder on. Switching to other question now....

12. Apr 10, 2015

### Fredrik

Staff Emeritus
I think the best way to make sense of the term "infinitesimal" in a physics book is to interpret it as a code that lets the experienced reader know that the next calculation will involve a Taylor series expansion and that only terms up to some specific order will be included.

13. Apr 11, 2015

### ognik

Thats kind of where I was, but it is sometimes unsatisfying ....The proof above doesn't work without it being 'neglected', so what does that say about T (and H) if we didn't have ε?

BTW, thanks for the encouragement BvU :-)

14. Apr 11, 2015

### Fredrik

Staff Emeritus
No, it does work. It's not about neglecting anything. It's about matching the coefficients of $\varepsilon^0,\varepsilon^1,\varepsilon^2,\dots$ on both sides. The fact that something had already been neglected in the calculation (actually in the definition of T) made it impossible to obtain a correct result from the coefficients of $\varepsilon^2$, but it had no effect on the coefficients on $\varepsilon^1$ or the coefficients of $\varepsilon^0$. We could neglect higher order terms because the result we're interested in is the equality of the coefficients of $\varepsilon$, but doing so had no advantages other than shortening the calculation.

15. Apr 11, 2015

### ognik

that makes good sense thanks - we end up equating the coefficients of all sorts of things.

Any thoughts on the other problem after my latest post? Just let me know if you need anything else ...

16. Apr 12, 2015

### Fredrik

Staff Emeritus
I took a look at it yesterday. I think I understand all the notations and stuff, but I wasn't able to prove the desired result for $\frac{\partial\hat{\mathbf q}_i}{\partial q_j}$ when $i\neq j$. I will try again later today.